Physics 11
HW #12 Solutions
Chapter 31:
Focus On Concepts: 6, 7, 9, 16
Problems: 3, 7, 25, 27, 28, 35, 39, 53, 56
Focus On Concepts 31-6
(d) The mass defect depends is equal to the mass of the separated nucleons (63 u) minus
the mass (62.5 u) of the stable nucleus (see Section 31.3).
Focus On Concepts 31-7
(a) Both α and β − decays produce a daughter nucleus that has a different atomic number
than the parent nucleus (see Section 31.4). Thus, each decay results in a new element.
Focus On Concepts 31-9
(d) The radius of a nucleus depends on the number of nucleons it contains (see Section
31.1). In α decay the number of nucleons in the daughter nucleus is four less than that in
the parent nucleus, while in β − and γ decay the number of nucleons is the same as that in
the parent nucleus (see Section 31.4).
Focus On Concepts 31-16
(d) The half-life is the time required for one-half of the original nuclei to disintegrate
(see Section 31.6).
Problem 31-3
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
A
X where A represents the number of protons and neutrons (the nucleon number) and Z
Z
represents the number of protons (the atomic number) in the nucleus.
SOLUTION
a. The symbol
195
78
X indicates that the nucleus in question contains Z = 78 protons, and
N = A − Z = 195 – 78 = 117 neutrons .
From the periodic table, we see that Z = 78
corresponds to platinum, Pt .
b. Similar reasoning indicates that the nucleus in question is sulfur, S , and the nucleus
contains N = A − Z = 32 – 16 = 16 neutrons .
c. Similar reasoning indicates that the nucleus in question is copper, Cu , and the nucleus
contains N = A − Z = 63 – 29 = 34 neutrons .
d. Similar reasoning indicates that the nucleus in question is boron, B , and the nucleus
contains N = A − Z = 11 – 5 = 6 neutrons .
plutonium, Pu , and the
e. Similar reasoning indicates that the nucleus in question is
nucleus contains N = A − Z = 239 – 94 = 145 neutrons .
Problem 31-7
REASONING The nucleus is roughly spherical, so its volume is V = 43 π r 3 . The radius r is
given by Equation 31.2 as r ≈ (1.2 × 10–15 m)A1/3, where A is the atomic mass number or
c
nucleon number. Therefore, the volume is given by V = 43 π 1.2 × 10 −15 m
h
3
A and is
proportional to A. We can apply this expression to the unknown nucleus and to the nickel
nucleus, knowing that the ratio of the two volumes is 2:1. This ratio provides the solution
we seek.
SOLUTION Applying the expression for the volume to each nucleus gives
c
h
3
4
π 1.2 × 10 −15 m A
V
A
3
=
=
=2
3
V Ni 4 π 1.2 × 10 −15 m 60
60
3
c
hb g
or
A = 120
The nucleon number is equal to the number of neutrons N plus the number of protons or the
atomic number Z, so A = N + Z. Therefore, the atomic number for the unknown nucleus is
Z = A – N = 120 – 70 = 50
The unknown nucleus, then, is
120
50
X . Reference to the periodic table reveals that the
element that has an atomic number of Z = 50 is tin (Sn). Thus,
A
Z
X= 120
Sn .
50
Problem 31-25
REASONING We can determine the identity of X in each of the decay processes by noting
that for each process, the sum of A and Z for the decay products must equal the values of A
and Z for the parent nuclei.
SOLUTION
a.
211
82
Pb → 211
Bi + X
83
Using the reasoning discussed above, X must have A = 0 and Z = –1 . Therefore X must
be an electron,
0
–1
e;
X represents a β – particle (electron) .
11
6
b.
C → 115 B + X
Similar reasoning suggests that X must have A = 0 and Z = +1 . Therefore X must be a
positron;
0
+1
e;
X represents a β + particle (positron) .
231
90
c.
Th * → 231
Th + X
90
Similar reasoning suggests that X must have A = 0 and Z = 0 . Therefore X must be a
gamma ray; X represents a γ ray .
210
84
d.
Po → 206
Pb + X
82
Using the reasoning discussed above, X must have A = 4 and Z = 2 . Therefore X must be
a helium nucleus, 42 He ; X represents an α particle (helium nucleus) .
Problem 31-27
REASONING Since energy is released during the decay, the combined mass of the lead
206
210
82 Pb daughter nucleus and the α particle is less than the mass of the polonium 84 Po
parent nucleus. The difference in mass is equivalent to the energy released. Since the recoil
of the lead nucleus is being ignored and we are assuming that all the released energy goes
into the kinetic energy KE of the α particle, it follows that released energy = KE = 12 mv 2
(Equation 6.2). Thus, the speed of the α particle is v =
2 ( KE )
.
m
SOLUTION The decay reaction is
210
Po
84
209.982 848 u
→
206
Pb
82
205.974 440 u
+
4
He
2
4.002 603 u
The difference in the masses is
209.982 848 u − 205.974 440 u − 4.002 603 u = 5.805 × 10−3 u
This mass difference corresponds to an energy of
(5.805 ×10−3 u ) ⎛⎜⎝ 931.51 uMeV ⎞⎟⎠ = 5.407 MeV
This energy is the kinetic energy of the α particle (mass m = 6.6447 × 10–27 kg, see Example
2), so the speed of the α particle is
v =
2 ( KE )
=
m
⎛ 1.60 × 10−19
2 5.407 × 106 eV ⎜
1 eV
⎝
−27
6.6447 × 10
kg
(
)
J⎞
⎟
⎠ = 1.61 × 107 m/s
Problem 31-28
REASONING When a nucleus undergoes an α decay, it emits one α particle that contains
two protons and two neutrons. Therefore, each α decay decreases the atomic mass number A
by four. A β − decay does not affect the atomic mass number A at all, so the difference
Ad − Ap between the atomic mass number Ad = 208 of the daughter nucleus
atomic mass number Ap = 230 of the parent nucleus
( 20882 Pb ) and the
( 23086 Rn ) after the emission of Nα alpha
particles is
Ad − Ap = −4 Nα
(1)
The negative sign in Equation (1) indicates a reduction in the atomic mass number in going
from the parent to the daughter. In order to determine the difference Zd − Zp between the
final atomic number Zd = 82 of the daughter nucleus and the atomic number Zp = 86 of the
parent nucleus, we must take into account both the loss of two protons for each α particle
emitted, and the gain of one proton for each β − particle emitted. Therefore, we have that
Zd − Z p = −2 Nα + N β
where Nβ is the number of β − particles emitted in the decay series.
SOLUTION Solving Equation (1) for Nα, we obtain
Nα =
Ad − Ap
−4
=
208 − 220
=3
−4
Solving Equation (2) for Nβ yields
N β = 2 Nα + Zd − Z p = 2(3) + 82 − 86 = 2
Therefore, in this decay series, 3 α particles and 2 β − particles are emitted.
(2)
Problem 31-35
REASONING AND SOLUTION The amount remaining is 0.0100% = 0.000 100. We know
N / N 0 = e –0.693t /T1/2 . Therefore, we find
t = −
⎛ N ⎞
29.1 yr
ln ⎜
ln(0.000 100) = 387 yr
⎟⎟ = −
⎜
0.693 ⎝ N 0 ⎠
0.693
T1/ 2
Problem 31-39
REASONING We can find the decay constant from Equation 31.5, N = N 0e – λt . If we
multiply both sides by the decay constant λ , we have
λ N = λ N 0e – λ t
A = A0e – λ t
or
where A0 is the initial activity and A is the activity after a time t. Once the decay constant is
known, we can use the same expression to determine the activity after a total of six days.
SOLUTION Solving the expression above for the decay constant λ , we have
λ = –
⎛ A⎞
⎛ 285 disintegrations/min ⎞
1
1
–1
ln ⎜ ⎟ = –
ln ⎜
⎟ = 0.167 days
⎜
⎟
t
2 days
⎝ 398 disintegrations/min ⎠
⎝ A0 ⎠
Then the activity four days after the second day is
A = (285 disintegrations/min) e – (0.167 days
–1
)(4.00 days)
= 146 disintegrations/min
Problem 31-53
REASONING The atomic mass given for 206
Pb includes the 82 electrons in the neutral
82
atom. Therefore, when computing the mass defect, we must account for these electrons. We
do so by using the atomic mass of 1.007 825 u for the hydrogen atom 11 H , which also
includes the single electron, instead of the atomic mass of a proton. To obtain the binding
energy in MeV, we will use the fact that 1 u is equivalent to 931.5 MeV.
SOLUTION
a. Noting that the number of neutrons is 206 – 82 = 124, we can obtain the mass defect Δm
as follows:
b
g
b
g
Δ m = 82 1.007 825 u + 124 1.008 665 u − 205.974 440 u = 1.741 670 u
82 free hydrogen atoms
(protons plus electrons)
124 free neutrons
Intact lead atom
(including 82 electrons)
b. Since 1 u is equivalent to 931.5 MeV, the binding energy is
b
Binding energy = 1.741 670 u
gFGH 931.15 uMeV IJK =
1622 MeV
c. The binding energy per nucleon is
Binding energy
Number of nucleons
Binding energy per nucleon =
1622 MeV
= 7.87 MeV/nucleon
206
=
Problem 31-56
REASONING AND SOLUTION
a. The decay reaction is: 242
Pu → ZA X + 42 He . Therefore, 242 = A + 4, so that A = 238. In
94
addition, 94 = Z + 2, so that Z = 92. Thus, the daughter nucleus is
b. The decay reaction is
24
11
13
7
U .
Na → ZA X + –10 e . Therefore, 24 = A. In addition, 11 = Z – 1, so
that Z = 12. Thus the daughter nucleus is
c. The decay reaction is
238
92
24
12
Mg .
N → ZA X + 01 e . Therefore, 13 = A. In addition, 7 = Z + 1, so that
Z = 6. Thus, the daughter nucleus is
13
6
C